浅谈Hex编码算法

 1 /**
 2      * 将字节数组转换为字符串
 3      * 一个字节会形成两个字符，最终长度是原始数据的2倍
 4      * @param data
 5      * @return
 6      */
 7     public static String toHex(byte[] data){
 8         String ret = null;
 9 
10         //TODO 将字节数组转换为字符串
11         if (data != null && data.length>0) {
12             StringBuilder sb = new StringBuilder();
13             for (byte b: data){
14                 //分别获取高四位，低四位的内容，将两个数值，转为字符
15                 int h = (b>>4)&0x0f;
16                 int l = b&0x0f;
17                 char ch ,cl;
18                 if( h > 9 ){
19                     ch = (char) (‘A‘+(h-10));
20                 }else{  //0--9
21                     ch = (char) (‘0‘+h);
22                 }
23 
24                 if(l>9){
25                     cl = (char) (‘A‘+(l-10));
26                 }else{  //0--9
27                     cl = (char) (‘0‘+l);
28                 }
29 
30 
31                     sb.append(ch).append(cl);
32                 }
33                   ret = sb.toString();
34             }
35 
36         return ret;
37     }

 1  public static byte[] fromHex(String str) {
 2         byte[] ret = null;
 3 
 4         //TODO 将Hex编码的字符串，还原为 原始的字节数组
 5         if (str != null) {
 6             int len = str.length();
 7             if (len > 0 && len % 2 == 0) {
 8                 char[] chs = str.toCharArray();
 9                 ret = new byte[len / 2];
10                 for (int i = 0, j = 0; i < len; i += 2, j++) {
11                     char ch = chs[i];
12                     char cl = chs[i + 1];
13 
14                     int ih = 0, il = 0, v = 0;
15                     if (ch >= ‘A‘ && ch <= ‘F‘) {
16                         ih = 10 + (ch - ‘A‘);
17                     } else if (ch >= ‘a‘ && ch <= ‘f‘) {
18                         ih = 10 + (ch - ‘a‘);
19                     } else if (ch >= ‘0‘ && ch <= ‘9‘) {
20                         ih = ch - ‘0‘;
21                     }
22 
23                     if (cl >= ‘A‘ && cl <= ‘F‘) {
24                         il = 10 + (cl - ‘A‘);
25                     } else if (cl >= ‘a‘ && cl <= ‘f‘) {
26                         il = 10 + (cl - ‘a‘);
27                     } else if (cl >= ‘0‘ && cl <= ‘9‘) {
28                         il = cl - ‘0‘;
29                     }
30 
31                     v = ((ih & 0x0f) << 4) | (il & 0x0f);
32                     //赋值
33                     ret[j] = (byte) v;
34                 }
35             }
36         }
37         return ret;
38     }